What if you had an algebraic equation like
that required two operations (in this case subtraction and then division) to solve it? After completing this Concept, you'll be able to solve two-step equations like this one.

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Guidance

We’ve seen how to solve for an unknown by isolating it on one side of an equation and then evaluating the other side. Now we’ll see how to solve equations where the variable takes more than one step to isolate.

Example A

Rebecca has three bags containing the same number of marbles, plus two marbles left over. She places them on one side of a balance. Chris, who has more marbles than Rebecca, adds marbles to the other side of the balance. He finds that with 29 marbles, the scales balance. How many marbles are in each bag? Assume the bags weigh nothing.

Solution

We know that the system balances, so the weights on each side must be equal. If we use
to represent the number of marbles in each bag, then we can see that on the left side of the scale we have three bags (each containing
marbles) plus two extra marbles, and on the right side of the scale we have 29 marbles. The balancing of the scales is similar to the balancing of the following equation.

“Three bags plus two marbles
equals
29 marbles”

To solve for
, we need to first get all the variables (terms containing an
) alone on one side of the equation. We’ve already got all the
’s on one side; now we just need to isolate them.

There are nine marbles in each bag.

We can do the same with the real objects as we did with the equation. Just as we subtracted 2 from both sides of the equals sign, we could remove two marbles from each side of the scale. Because we removed the same number of marbles from each side, we know the scales will still balance.

Then, because there are three bags of marbles on the left-hand side of the scale, we can divide the marbles on the right-hand side into three equal piles. You can see that there are nine marbles in each.

Three bags of marbles
balances
three piles of nine marbles.

So each bag of marbles balances nine marbles, meaning that each bag contains nine marbles.

Example B

Solve
.

This equation has the
buried in parentheses. To dig it out, we can proceed in one of two ways: we can either distribute the six on the left, or divide both sides by six to remove it from the left. Since the right-hand side of the equation is a multiple of six, it makes sense to divide. That gives us
. Then we can subtract 4 from both sides to get
.

Example C

Solve
.

It’s always a good idea to get rid of fractions first. Multiplying both sides by 5 gives us
, and then we can add 3 to both sides to get
.